The weights of 5 students are x, x+5, 2x-40, 65, and 70 kg. The mean weight is 60 kg. Find x, list the weights, and calculate the median and standard deviation.

The weights of the 5 students are $x$, $x+5$, $2x-40$, 65, and 70 kg.
The mean weight is given as 60 kg.
The formula for the mean is the sum of all values divided by the number of values.

$$\text{Mean} = \frac{\sum \text{weights}}{\text{Number of students}}$$ $$60 = \frac{x + (x+5) + (2x-40) + 65 + 70}{5}$$

Multiply both sides by 5:

$$60 \times 5 = x + x + 5 + 2x - 40 + 65 + 70$$ $$300 = (x+x+2x) + (5 - 40 + 65 + 70)$$ $$300 = 4x + 100$$

Subtract 100 from both sides:

$$300 - 100 = 4x$$ $$200 = 4x$$

Divide by 4:

$$x = \frac{200}{4}$$ $$x = 50$$

Substitute $x=50$ into the expressions for the weights:

• First weight: $x = 50$ kg
• Second weight: $x+5 = 50+5 = 55$ kg
• Third weight: $2x-40 = 2(50)-40 = 100-40 = 60$ kg
• Fourth weight: 65 kg
• Fifth weight: 70 kg The weights are $50, 55, 60, 65, 70$ kg.

i) Calculate the median.
To find the median, arrange the weights in ascending order. The weights are already in order: $50, 55, 60, 65, 70$.
Since there are 5 data points (an odd number), the median is the middle value.
The position of the median is $\frac{n+1}{2}$, where $n$ is the number of data points.
Position = $\frac{5+1}{2} = \frac{6}{2} = 3^\text{rd}$ value.
The 3rd value in the ordered list is 60.
The median is $\boxed{\text{60 kg}}$.

ii) Calculate the standard deviation.
The formula for the population standard deviation ($\sigma$) is:

$$\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}$$

where $x_i$ are the individual weights, $\mu$ is the mean, and $N$ is the number of students.
We have $N=5$ and $\mu=60$ kg.

Calculate the squared difference from the mean for each weight:

• $(50 - 60)^2 = (-10)^2 = 100$
• $(55 - 60)^2 = (-5)^2 = 25$
• $(60 - 60)^2 = (0)^2 = 0$
• $(65 - 60)^2 = (5)^2 = 25$
• $(70 - 60)^2 = (10)^2 = 100$

Sum of the squared differences:

$$\sum (x_i - \mu)^2 = 100 + 25 + 0 + 25 + 100 = 250$$

Now, substitute this into the standard deviation formula:

$$\sigma = \sqrt{\frac{250}{5}}$$ $$\sigma = \sqrt{50}$$ $$\sigma = \sqrt{25 \times 2}$$ $$\sigma = 5\sqrt{2}$$

As a decimal, $\sigma \approx 7.071$ kg.
The standard deviation is $\boxed{\text{7.071 kg (or } 5\sqrt{2} \text{ kg)}}$.